a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)

\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

hay \(x=-\dfrac{1}{4}\)

c) Ta có: \(8x^3-50x=0\)

\(\Leftrightarrow2x\left(4x^2-25\right)=0\)

\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

\(=\left(x^3-2x^2+x+2x^2-4x+2-2x+7\right):\left(x^2-2x+1\right)\\ =\left[\left(x^2-2x+1\right)\left(x+2\right)-2x+7\right]:\left(x^2-2x+1\right)\\ =x+2\left(dư:-2x+7\right)\)

Mình không viết lại đề nhé

a) -12x + 60 + 21 - 7x = 5

-19x = 5 - 71

-19x = -76

x = 4

b) 3 - 17 + x = 289 - 36 - 289

x = -22

\(a,-12.\left(x-5\right)+7.\left(3-x\right)=5\)

\(=.-12x+60+21-7x=5\)

\(=>-19x=5-60-21=-76\)

\(=>x=\frac{-76}{-19}=\frac{76}{19}=4\)

\(b,3-\left(17-x\right)=289-\left(36+289\right)\)

\(=>3-17+x=-36\)

\(=>x=-36+17-3=-22\)

a) -12(x-5)+7(3-x)=5

<=> -12x+60+21-7x=5

<=> 9+53x=5

<=> 53x=-4

<=> x=\(\frac{-4}{53}\)

b) 3-(17-x)=289-(36+289)

<=> 3-17+x=289-36-289

<=> -14+x=-36

<=> x=-22

khó quá nhiw

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

A.  2.\(|3x+1|\)=\(\frac{3}{4}\)-\(\frac{5}{8}\)

     2.\(|3x+1|\)=1/8

        \(|3x+1|\)=1/8:2

        \(|3x+1|\)=1/16

TH1 : 3x+1=1/16

         3x=1/16-1

         3x=-15/16

         x=-15/16:3

          x=-5/16

a,\(\frac{3}{4}-2.\left|3x+1\right|=\frac{5}{8}\)

\(\Rightarrow2.\left|3x+1\right|=\frac{3}{4}-\frac{5}{8}=\frac{6}{8}-\frac{5}{8}=\frac{1}{8}\)

\(\Rightarrow\left|3x+1\right|=\frac{1}{8}.\frac{1}{2}=\frac{1}{16}\)

\(\Rightarrow\orbr{\begin{cases}3x+1=\frac{1}{16}\\3x+1=\frac{-1}{16}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{16}-1=\frac{-15}{16}\\3x=\frac{-1}{16}-1=\frac{-17}{16}\end{cases}}\)

                                          \(\Rightarrow\orbr{\begin{cases}x=\frac{-15}{16}.\frac{1}{3}=\frac{-5}{16}\\x=\frac{-17}{16}.\frac{1}{3}=\frac{-17}{48}\end{cases}}\)

Vậy....

b,\(\left|3x+2\right|-\left|x-3\right|=\frac{7}{2}\left(1\right)\)

Ta có bảng xét dấu

Nếu x<\(\frac{-2}{3}\)       thì \(\left|3x+2\right|-\left|x-3\right|\) \(=-3x-2-3+x\)

                                                                         \(=-2x-5\)

Từ (1) \(\Rightarrow-2x-5=\frac{7}{2}\)

          \(\Rightarrow-2x=\frac{7}{2}+5=\frac{17}{2}\)

           \(\Rightarrow x=\frac{17}{2}\cdot\frac{-1}{2}=\frac{-17}{4}\)(thỏa mãn x<\(\frac{-2}{3}\)

Nếu \(\frac{-2}{3}\le x\le3\)thì \(\left|3x+2\right|-\left|x-3\right|=3x+2-\left(3-x\right)\)

                                                                                \(=3x+2-3+x\)

                                                                                 \(=2x-1\)

Từ (1)\(\Rightarrow\)\(2x-1=\frac{7}{2}\)

    \(\Rightarrow2x=\frac{9}{2}\)

      \(\Rightarrow x=\frac{9}{4}\)(thỏa mãn......

Còn trưonwfg hợp cuối bạn tự làm nốt nhé

a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)

=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)

b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)

=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)

TThế giới oi oi oi 

a: \(\dfrac{x}{6}=\dfrac{8}{3}\)

=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)

b: \(\dfrac{5}{x}=\dfrac{4}{9}\)

=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)

c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)

=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)

=>x=-1-3=-4

d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)

=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)

=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)

=>\(x=-\dfrac{69}{8}\)

f: ĐKXĐ: x<>1

\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)

=>\(\left(x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)